Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int n, m;
const long long int N = 105;
long long int arr[N][N];
long long int sarr[N][N];
long long int rowmin[N];
long long int colmin[N];
vector<pair<string, long long int> > v;
bool check() {
  long long int flag = 0;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++)
      if (sarr[i][j] != arr[i][j]) flag = 1;
  }
  if (flag)
    return false;
  else
    return true;
}
long long int rowover() {
  long long int ans = 0;
  long long int inf = 1e18;
  for (long long int i = 1; i <= n; i++) rowmin[i] = inf;
  for (long long int i = 1; i <= m; i++) colmin[i] = inf;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++)
      rowmin[i] = min(rowmin[i], arr[i][j]);
  }
  for (long long int i = 1; i <= m; i++) {
    for (long long int j = 1; j <= n; j++)
      colmin[i] = min(colmin[i], arr[j][i] - rowmin[j]);
  }
  memset(sarr, 0, sizeof(sarr));
  for (long long int i = 1; i <= n; i++) {
    sarr[i][1] += rowmin[i];
    for (long long int j = 1; j <= m; j++) sarr[i][j] += sarr[i][j - 1];
  }
  for (long long int i = 1; i <= m; i++) {
    for (long long int j = 1; j <= n; j++) sarr[j][i] += colmin[i];
  }
  if (check()) {
    v.clear();
    for (long long int i = 1; i <= n; i++) {
      if (rowmin[i] > 0) {
        while (rowmin[i]--) v.push_back({"row", i});
      }
    }
    for (long long int i = 1; i <= m; i++) {
      if (colmin[i] > 0) {
        while (colmin[i]--) v.push_back({"col", i});
      }
    }
    ans = v.size();
    return ans;
  } else
    return -1;
}
long long int colover() {
  long long int ans = 0;
  long long int inf = 1e18;
  for (long long int i = 1; i <= n; i++) rowmin[i] = inf;
  for (long long int i = 1; i <= m; i++) colmin[i] = inf;
  for (long long int i = 1; i <= m; i++) {
    for (long long int j = 1; j <= n; j++)
      colmin[i] = min(colmin[i], arr[j][i]);
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++)
      rowmin[i] = min(rowmin[i], arr[i][j] - colmin[j]);
  }
  memset(sarr, 0, sizeof(sarr));
  for (long long int i = 1; i <= n; i++) {
    sarr[i][1] += rowmin[i];
    for (long long int j = 1; j <= m; j++) sarr[i][j] += sarr[i][j - 1];
  }
  for (long long int i = 1; i <= m; i++) {
    for (long long int j = 1; j <= n; j++) sarr[j][i] += colmin[i];
  }
  if (check()) {
    v.clear();
    for (long long int i = 1; i <= n; i++) {
      if (rowmin[i] > 0) {
        while (rowmin[i]--) v.push_back({"row", i});
      }
    }
    for (long long int i = 1; i <= m; i++) {
      if (colmin[i] > 0) {
        while (colmin[i]--) v.push_back({"col", i});
      }
    }
    ans = v.size();
    return ans;
  } else
    return -1;
}
void solve() {
  cin >> n >> m;
  memset(arr, 0, sizeof(arr));
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j <= m; j++) cin >> arr[i][j];
  }
  long long int ans1 = rowover();
  long long int ans2 = colover();
  if (ans1 == -1 && ans2 == -1)
    cout << -1 << endl;
  else {
    if (ans1 < ans2) {
      rowover();
    }
    cout << v.size() << endl;
    for (long long int i = 0; i < v.size(); i++)
      cout << v[i].first << " " << v[i].second << endl;
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}